x \displaystyle \left| x \right| , Hoplessly Positive.

Algebra Level 2

If x 2 + x 6 = 0 \displaystyle { \left| x \right| }^{ 2 }+\left| x \right| -6=0 ,

then which of the fowling option is true?

  • (A) Product of roots is 6 \displaystyle -6

  • (B) Sum of the roots is + 1 \displaystyle +1 or 1 \displaystyle -1

  • (C) Both (A) and (B)

  • (D) None of these.

AYWC?
(C) (D) (B) (A)

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Soumo Mukherjee
Jan 10, 2015

Note: x \displaystyle \left| x \right| isn't the variable. x \displaystyle x is the variable.


x 2 + x 6 = 0 ( x + 3 ) ( x 2 ) = 0 \displaystyle { \left| x \right| }^{ 2 }+\left| x \right| -6=0\\ \Rightarrow \left( \left| x \right| +3 \right) \left( \left| x \right| -2 \right) =0 .

But, x + 3 0 \displaystyle \left| x \right| +3\neq 0 because x 3 \displaystyle \left| x \right| \neq -3 (i.e x \left| x \right| cannot be negative.)

Therefore, x 2 = 0 x = 2 o r 2. \displaystyle \left| x \right| -2=0\\ \Rightarrow x=2\quad or\quad -2. .

Clearly sum of roots is 2 + ( 2 ) = 0 \displaystyle 2+\left( -2 \right) =0 and product of roots is 2 × ( 2 ) = 4 \displaystyle 2\times \left( -2 \right) =-4 . None of which are given in the options.

Hence the answer is (D) None of these ''.

Well actually there are also infinitely many complex roots, e.g. x = ± 2 i , x = \pm 2i, x = ( 1 ± i ) 2 , x = (1 \pm i)\sqrt{2}, etc.

Edit: the general solution is 2 Re ( x ) 2 , Im ( x ) = ± 4 ( Re ( x ) ) 2 -2\leq\text{Re}(x)\leq 2,\\ \text{Im}(x) = \pm\sqrt{4 - (\text{Re}(x))^2} It is a circle of radius 2 2 centered about the origin in the complex plane. So the product is divergent and the sum is 0. 0.

Caleb Townsend - 6 years, 2 months ago

or just observing the coefficents

Mardokay Mosazghi - 6 years, 4 months ago

Log in to reply

Sorry, what?

Soumo Mukherjee - 6 years, 4 months ago

Log in to reply

vieta formula

Mardokay Mosazghi - 6 years, 4 months ago

Log in to reply

@Mardokay Mosazghi hmm...we are having extraneous roots, will vieta work?

Soumo Mukherjee - 6 years, 4 months ago

Log in to reply

@Soumo Mukherjee i agree i guess it was just an observation btw nice solution sir

Mardokay Mosazghi - 6 years, 4 months ago

Log in to reply

@Mardokay Mosazghi Sir?

plz don't call me sir :D

Soumo Mukherjee - 6 years, 4 months ago

Log in to reply

@Soumo Mukherjee ok bro but sir is a term used for respect

Mardokay Mosazghi - 6 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...